Question

How do find the vertex and axis of symmetry for a quadratic equation `y=4x^2-2x+2`?

Answer 1

The general form of a parabolic equation ins vertex form is:
`color(white)("XXXX")` `y = m(x-a)^2+b` where the vertex is at `(a,b)`

`y=4x^2-2x+2`
can be converted into that form (most easily by a completion of the squares process)
`color(white)("XXXX")` `y = 4(x^2- 1/2x) +2`

`color(white)("XXXX")` `y= 4(x^2-1/2x+(1/4)^2) + 2 -4(1/4)^2`

`color(white)("XXXX")` `y = 4(x-1/4)^2 + 7/4`

The vertex is at `(1/4, 7/4)`